# Probability - density

1. Mar 24, 2010

### XodoX

1. The probability density function for the continuous random vector X = (X1, X2) is given by:

Fx1, x2(X1,X2) = { 1/2 if X1+X2$$\leq$$ 2,X1$$\geq$$ 0,X2$$\geq$$0
0 otherwise }

Calculate the probability that X1>2X2

I do not know which formular to use. I can't find the right one. How do I do this? The joint density formular?

2. Mar 24, 2010

### LCKurtz

Draw a picture in the x1x2 plane showing where the density is 1/2. Then, on the same picture, draw the region in question where x1>2x2. That should show you the region to work with.

3. Mar 24, 2010

### XodoX

Yeah, I did that. However, I don't know which formular to use.

4. Mar 24, 2010

### Dick

You integrate the probability density over the region where x1>2*x2.

5. Mar 24, 2010

### XodoX

So joint density was right?

6. Mar 24, 2010

### Dick

If that's what you call integrating probability density over part of your region where x1>2*x2, then sure.

7. Mar 24, 2010

### XodoX

Joint density would be:

F x,y(x,y)=xe-x(y+1), x>0 ,y=0

8. Mar 24, 2010

### Dick

That's not it. The region where density is 1/2 is a triangle. Does the picture you drew show that?

9. Mar 24, 2010

### XodoX

Well, I actually didn't know that this one can not be used for triangles?
There's another formular I found for joint density...

F N,X(n,x)=Xne-2x/n!

10. Mar 24, 2010

### Dick

Stop looking up formulas. You don't need them. It's not that complicated. LCKurtz said to draw a picture. You said you did. What does it look like? Describe the triangle where density is 1/2.

11. Mar 24, 2010

### LCKurtz

Huh?? Where did that come from? You gave us the joint density in the original post. And the probability that X1>2X2 should be a number between 0 and 1.

12. Mar 25, 2010

### XodoX

Well, I eventually solved it myself. Integration wasn't needed.

13. Mar 25, 2010

### Dick

Good. That's always the best way to solve it. And, yes, you don't need explicit integration since the probability density isn't a function of x1 and x2. You just find the overlap area and multiply by 1/2.

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